# The Unapologetic Mathematician

## Differentiating Partial Integrals

A neat little variation on differentiating an integral from last time combines it with the fundamental theorem of calculus. It’s especially interesting in the context of evaluating iterated integrals for irregular regions where the limits of integration may depend on other variables.

Let $f:R\rightarrow\mathbb{R}$ is a continuous function on the rectangle $[a,b]\times[c,d]$, and that $D_2f$ is also continuous on $R$. Also, let $p:[c,d]\rightarrow[a,b]$ and $q:[c,d]\rightarrow[a,b]$ be two differentiable functions on $[c,d]$ with images in $[a,b]$. Define the function

$\displaystyle F(y)=\int\limits_{p(y)}^{q(y)}f(x,y)\,dx$

Then the derivative $F'(y)$ exists and has the value

$\displaystyle F'(y)=\int\limits_{p(y)}^{q(y)}\left[D_2f\right](x,y)\,dx+f(q(y),y)q'(y)-f(p(y),y)p'(y)$

In fact, if we forget letting $f$ depend on $y$ at all, this is the source of some of my favorite questions on first-semester calculus finals.

Anyway, we define another function for the moment

$\displaystyle G(x^1,x^2,x^3)=\int\limits_{x^1}^{x^2}f(t,x^3)\,dt$ for $x^1$ and $x^2$ in $[a,b]$ and $x^3$ in $[c,d]$. Then $F(y)=G(p(y),q(y),y)$.

The fundamental theorem of calculus tells us the first two partial derivatives of $G$ immediately, and for the third we can differentiate under the integral sign:

\displaystyle\begin{aligned}\frac{\partial G}{\partial x^1}&=-f(x^1,x^3)\\\frac{\partial G}{\partial x^2}&=f(x^2,x^3)\\\frac{\partial G}{\partial x^3}&=\int\limits_{x^1}^{x^2}\left[D_2f\right](t,x^3)\,dt\end{aligned}

Then we can use the chain rule:

\displaystyle\begin{aligned}\frac{d}{dy}F(y)&=\frac{d}{dy}G(p(y),q(y),y)\\&=\frac{\partial G}{\partial x^1}\frac{dp}{dy}+\frac{\partial G}{\partial x^2}\frac{dq}{dy}+\frac{\partial G}{\partial x^3}\frac{dy}{dy}\\&=-f(p(y),y)p'(y)+f(q(y),y)q'(y)+\int\limits_{p(y)}^{q(y)}\left[D_2f\right](x,y)\,dx\end{aligned}

January 14, 2010 - Posted by | Analysis, Calculus

1. Please allow me to write here an easy exercise (in a different notation of yours).

$I(t)=J(u,v,t)=\displaystyle\int_{u(t)}^{v(t)}f\left( x,t\right) dx$

$I^{\prime }(t)=\displaystyle\int_{u\left( t\right) }^{v\left( t\right) }\dfrac{\partial f\left( x,t\right) }{\partial t}dx+f\left( v\left( t\right) ,t\right) v^{\prime }\left( t\right) -f\left( u\left( t\right) ,t\right) u^{\prime}\left( t\right)$.

Exercise. Find the derivative $I^{\prime }(t)$ of the integral

$I(t)=J(2t,t^{2},t)=\displaystyle\int_{2t}^{t^{2}}e^{tx}dx$

Solution. In this case we have $f\left( x,t\right) =e^{tx},u\left( t\right) =2t$ and $v\left( t\right) =t^{2}.$ The derivatives are

$v^{\prime }\left( t\right) =2t\qquad\qquad u^{\prime }\left( t\right) =2$

$\dfrac{\partial f\left( x,t\right) }{\partial t}=\dfrac{\partial }{\partial t}e^{tx}=xe^{tx}$.

The values of the integrand function are evaluated at $\left( v,t\right)$ and $\left( u,t\right)$

$f\left( v\left( t\right) ,t\right) =e^{t\cdot t^{2}}=e^{t^{3}}$

$f\left( u\left( t\right) ,t\right) =e^{t\cdot 2t}=e^{2t^{2}}$

Hence

$I^{\prime }(t)=\displaystyle\int_{u\left( t\right) }^{v\left( t\right) }\dfrac{\partial f\left( x,t\right) }{\partial t}dx+f\left( v\left( t\right) ,t\right) v^{\prime }\left( t\right) -f\left( u\left( t\right) ,t\right) u^{\prime }\left( t\right)$ $=\displaystyle\int_{2t}^{t^{2}}xe^{tx}dx+2te^{t^{3}}-2e^{2t^{2}}$
$=\dfrac{e^{t^{3}}\left( 3t^{3}-1\right) -e^{2t^{2}}\left( 4t^{2}-1\right) }{t^{2}}$

Comment by Américo Tavares | January 15, 2010 | Reply

2. You keep writing out these “exercises” in the comments.. why not make posts about them and link here? I know you have your own weblog, and you aren’t really commenting at all.

Comment by John Armstrong | January 15, 2010 | Reply

3. I wrote them here because I thought they were useful and appropriate. Since that is not the case, I will stop writing them as comments here. Of course you can delete them all.

Comment by Américo Tavares | January 15, 2010 | Reply